Question C asks how's to calculate the current over the led and 3k resistor? Can someone please help me understand how to do it? enter image description here

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    \$\begingroup\$ I(3k)=I(470Ω)+I(LED) and Vled=2V so V(3k)=18V got it? \$\endgroup\$ Dec 18 '18 at 22:39
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    \$\begingroup\$ In the future please use a photo editor like GIMP or Photoshop to remove the stuff we do not need to see. You can use the cropping tool to select just the section we need to see. \$\endgroup\$
    – user105652
    Dec 18 '18 at 23:01
  • \$\begingroup\$ Convert the original voltage source and the resistor divider into a new Thevenin voltage source and Thevenin output resistance. Subtract the diode voltage from the Thevenin voltage source value, then divide this resulting voltage by the Thevenin resistance you computed to get the LED current. \$\endgroup\$
    – jonk
    Dec 18 '18 at 23:43

enter image description here

20V supply and 2V across the LED leaves how much voltage across the 3K Ω resistor?

With that voltage, use ohms law to calculate the current through the 3K Ω resistor.

Calculate the current for a 470 Ω resistor at 2v.
There is more current flowing through the 3K Ω than the 470 Ω resistor.
Where would that excess I3K - I470Ω current go?

  • \$\begingroup\$ I'll upvote since the OP did not. \$\endgroup\$
    – user103380
    Dec 19 '18 at 16:28
  • \$\begingroup\$ Ye I've basically done that myself. I just wasn't sure what I was doing was right. Appreciate the help! Also give me chance to reply KingDuken. \$\endgroup\$
    – ThePandeo
    Dec 19 '18 at 19:21

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